Báo cáo toán học: "Extremal problems for independent set enumeration" ppsx

If we instead restrict the class of graphs to those on a fixed number of vertices and edges, then the Kruskal-Katona theorem implies that the graph with the maximum number of independent

Extremal problems for independent set enumeration

Jonathan Cutler

Montclair State University New Jersey, U.S.A

jonathan.cutler@montclair.edu

A.J Radcliffe

University of Nebraska-Lincoln Nebraska, U.S.A

aradcliffe1@math.unl.edu Submitted: Jun 20, 2011; Accepted: Aug 16, 2011; Published: Aug 26, 2011

Mathematics Subject Classification: 05C35, 05C69

Abstract The study of the number of independent sets in a graph has a rich history Recently, Kahn proved that disjoint unions of Kr,r’s have the maximum number

of independent sets amongst r-regular bipartite graphs Zhao extended this to all r-regular graphs If we instead restrict the class of graphs to those on a fixed number of vertices and edges, then the Kruskal-Katona theorem implies that the graph with the maximum number of independent sets is the lex graph, where edges form an initial segment of the lexicographic ordering In this paper, we study three related questions Firstly, we prove that the lex graph has the maximum number

of weighted independent sets for any appropriate weighting Secondly, we solve the problem of maximizing the number of independents sets in graphs with specified independence number or clique number Finally, for m ≤ n, we find the graphs with the minimum number of independent sets for graphs with n vertices and m edges

1 Introduction

The study of independent sets in graphs has a long and rich history We write I(G) for the set of independent sets in a graph G and i(G) = |I(G)| for the number of independent sets Much recent research has focused on the problem of maximizing the number of independent sets in graphs with certain restrictions For example, Kahn [9] gave the following upper bound for regular bipartite graphs

Theorem 1.1 (Kahn) Let G be an r-regular bipartite graph on n vertices Then

i(G) ≤ 2r+1− 1
n/2r

The result is sharp for disjoint unions of copies of Kr,r Zhao [19] proved a conjecture

of Kahn by extending Theorem 1.1 to all r-regular graphs Kahn was also interested in

weighted independent sets Assign a weight of λk (for some λ > 0) to an independent set of size k and set iλ(G) to be the total weight of all independent sets in G Kahn also showed that if G is an r-regular bipartite graph on n vertices, and λ ≥ 1, then

iλ(G) ≤ iλ(Kr,r)n/2r Galvin and Tetali [7] extended this result to all positive weights

In this paper, we will be concerned with the number of independent sets in graphs from a variety of graph classes As noted in [4], one can prove, as a consequence of the Kruskal-Katona theorem [11, 10], that the graph on n vertices having m edges that has the largest number of independent sets is the lex graph, denoted L(n, m), which we define

in the next section

Theorem 1.2 If G is a graph on n vertices and m edges, then

i(G) ≤ i(L(n, m))

This result was also proved by Wood [17], though he was working in the framework of counting cliques, i.e., complete subgraphs

This paper will focus on three independent set enumeration problems In Section 2,

we extend Theorem 1.2 to the enumeration of independent sets of fixed size and deduce a theorem about weighted independent set enumeration We also note some consequences for homomorphism enumeration In Section 3, we extend our study to classes of graphs determined by bounds on their independence number and clique number In Section 4,

we will study the problem of minimizing the number of independent sets among graphs

of fixed order and size

2 Consequences of the Kruskal-Katona theorem

The Kruskal-Katona theorem [11, 10] has many consequences for the study of independent sets in graphs We summarize some of them in this section, after introducing the lex orderings

Definition 1 The lexicographic (or lex ) ordering on subsets of N is defined as follows Given A, B ⊆ N we say A precedes B in lex, written A ≺LB, if min(A 4 B) ∈ A Further, we define the lex graph with n vertices and m edges, denoted L(n, m), to

be the graph with vertex set [n] and edge set the first m elements of [n]2
under the lex ordering The first few elements of the lex order on [n]2
are

{1, 2} , {1, 3} , {1, 4} , , {1, n} , {2, 3} , {2, 4} , , {2, n} , {3, 4} , ,

and so the lex graph evolves by successively saturating the vertices 1, 2, 3, , n in order Our concern will mostly be with the lex ordering restricted to the level sets [n]k
The Kruskal-Katona theorem says that among all subsets of [n]k
of given size the one with the

smallest “upper shadow” on levels above k is the initial segment of lex1 The following definition makes the notion of shadow precise

Definition 2 If A ⊆ [n]k
and ` ≥ k then the upper `-shadow of A is the collection

∂`(A) =



B ∈[n]

`

 : for some A ∈ A we have B ⊇ A



Theorem 2.1 (Kruskal-Katona) If A ⊆ [n]k
has size a and L(a) ⊆ [n]

k
is the initial segment of size a in the lex order on [n]k
then for all ` ≤ k we have

∂`(A)≥

∂`(L(a))

We derive below some consequences of Theorem 2.1 for extremal problems in indepen-dent set enumeration The study of indepenindepen-dent sets of a fixed size has generated interest when the graphs considered are regular, but seem to have not been well-studied if the graphs have a fixed number of edges Carroll, Galvin, and Tetali [2] gave asymptotics on the maximum number of independent sets of a fixed size in regular graphs Theorem 2.2 shows that the lex graph maximizes the number of independent sets of a fixed size, and so also maximizes weighted independent set counts for arbitrary weight functions depending only on the size of the set

Definition 3 If w : N ∪ {0} → [0, ∞) is any function and G is a graph then we set

iw(G) = X

I∈I(G)

w(|I|)

If k ∈ N ∪ {0} then in particular we define wk by wk(i) = 1(i = k), and

ik(G) = iwk(G) = |{I ∈ I(G) : |I| = k}| Theorem 2.2 Let G be a graph on n vertices with m edges If w : N ∪ {0} → [0, ∞) is any function then

iw(G) ≤ iw(L(n, m))

Proof It clearly suffices to prove the special case where w = wkfor some k ∈ N, since any other w can be expressed as a positive linear combination of these We may also assume that V (G) = [n] Now let Ik = nA ∈ [n]k

: A is independent in Go and ik = |Ik| We have I ∈ Ik iff I ∈ [n]k
\ ∂k(E(G)), so

ik =n

k



− ∂k(E(G))≤n

k



− ∂k(L(n, m))= ik(L(n, m))

1 The Kruskal-Katona theorem is more usually stated in terms of lower shadows, and initial segments

in the colex order, defined for finite subsets of N by A ≺ C B if max(A 4 B) ∈ B The two forms are equivalent by a double reversal: complementation of sets and reversing the ordering on [n].

A natural set of weight functions arises by considering independent set enumeration as

a special case of homomorphism enumeration For graphs G and H, let Hom(G, H) be the set of homomorphisms from G to H and hom(G, H) = |Hom(G, H)| Letting HI be the path on two vertices, with one vertex looped and the other unlooped (and vertices labeled

as in Figure 1), we see that i(G) = hom(G, HI) This is because there is a one-to-one correspondence between independent sets A and homomorphisms φ, given by φ−1({a}) =

A In the context of statistical mechanics, it is natural to weight homomorphisms in the

Figure 1: The graph HI

following manner: each vertex x in H is assigned a weight β(x) Then we “count” the number of homomorphisms from G to a graph H with weight function β : V (H) → [0, ∞)

as follows:

homβ(G, H) = X

φ∈Hom(G,H)

Y

v∈G

β(φ(v))

This is the partition function of G for the model specified by H Note that if β ≡ 1, then homβ(G, H) = hom(G, H) Let βλ be the weight function on HI defined as follows

βλ(x) =

(

λ if x = a

1 if x = b

Finally, note that iλ(G) = homβλ(G, HI) Note that an independent set of size k is assigned weight λk Thus, the following is a corollary of Theorem 2.2

Corollary 2.3 If G is a graph on n vertices and m edges and λ > 0, then

iλ(G) ≤ iλ(L(n, m))

For more general image graphs H, the problem of finding the graph G with a given number of vertices and edges that maximizes hom(G, H) is usually difficult In [4], the authors solved the case where H is the Widom-Rowlinson graph, a P3 with every vertex looped In [5], the similar case where one end vertex of the P3 is unlooped is solved For one class of image graphs, it follows from Corollary 2.3 that the lex graph is an extremal graph Let S(p, q) be the clique-looped split graph Kp∨ Eq, in which each vertex of the

Kp is looped

Corollary 2.4 Let H = S(p, q) for p, q ≥ 1 and G be a graph with n vertices and m edges Then

hom(G, H) ≤ hom(L(n, m), H)

Proof Set λ = p/q Then, we have hom(G, H) = qniλ(G) and so the result follows from Corollary 2.3

3 Classes defined by other parameters

While independent sets in regular graphs and graphs of fixed size have been well studied, there has been less work on the problem of maximizing the number of independent sets in classes of graphs defined by other graph parameters In this section, we investigate classes defined by bounds on the independence number, α(G), and the clique number, ω(G) The first problem we attack is that of determining

max {i(G) : n(G) = n, α(G) ≤ α} Definition 4 If G is any graph and S is a subset of V (G), we let i(G; S) be the number

of independent sets in G containing S, and similarly ik(G; S) be the number of such independent sets of size k

We will prove the following theorem

Theorem 3.1 If G is a graph with n vertices and α(G) ≤ α, then

i(G) ≤ i(Tn,α), where Tn,α is the Tur´an graph with α parts That is,

i(G) ≤ i (Kn1 ∪ Kn2 ∪ · · · ∪ Knα) , where P ni = n and n1 ≤ n2 ≤ · · · ≤ nα ≤ n1+ 1

This follows from a more detailed result that among all graphs with independence number at most α, the complement of the Tur´an graph Tn,α has the maximum number

of independent sets of each fixed size This result is equivalent, by complementation, to the result of Zykov [20] which states that among all graphs with ω(G) ≤ α, the Tur´an graph Tn,α has the maximum number of cliques of each fixed size This result has been independently proved by many people, including Erd˝os [6], Sauer [15], Hadˇziivanov [8], and Roman [14] We give here a direct proof that is substantially shorter than earlier ones Our proof is based on the fifth proof of Tur´an’s theorem (the origin of which seems

to be unknown) in Proofs from the Book [1]

Theorem 3.2 If G is a graph with n vertices and α(G) ≤ α and 0 ≤ k ≤ n, then

ik(G) ≤ ik(Kn1 ∪ Kn2 ∪ · · · ∪ Knα) , where P ni = n and n1 ≤ n2 ≤ · · · ≤ nα ≤ n1+ 1

Proof Let G have n vertices, α(G) ≤ α, ik(G) maximal, and, subject to these conditions, e(G) maximal It suffices to show that G is of the form Km1∪ Km2 ∪ · · · ∪ Kmα for some

mi ≥ 0 since for such G,

ik(G) = X

A⊆([α]

k)

Y

j∈A

mj ≤ ik(Kn1 ∪ Kn2 ∪ · · · ∪ Knα)

We begin by showing that adjacency is an equivalence relation on V (G) The symmetry and reflexivity of ∼ are obvious, so transitivity is all that needs to be proved Suppose that u ∼ v ∼ w, but u 6∼ w We consider two cases depending on the relative sizes of

ik(G; u), ik(G; v), and ik(G; w)

If ik(G; u) > ik(G; v) (or ik(G; w) > ik(G; v)), then we change G by deleting v and then cloning u To be precise, we form G0 by deleting v from G and adding a new vertex

u0 where NG0(u0) = NG(u) ∪ {u} \ {w} No independent set in G0 contains both u and u0, since u ∼ u0, and so α(G) ≤ α Further,

ik(G) = ik(G) − ik(v) + ik(u) > ik(G), contradicting the maximality of G

On the other hand, if ik(G; v) ≥ ik(G; u), ik(G; w), then we consider the graph in which

we delete u and w and clone v twice That is, we form G00 from G − u − w by adding vertices v0 and v00 joined to each other, v, and NG(v) \ {u, w} Again, no independent set

in G00 contains any two of v, v0, and v00 since they are mutually adjacent, so α(G00) ≤ α

We have

ik(G00) = ik(G − u − w) + 2ik(G; v)

= ik(G) − ik(G; u) − ik(G; w) + ik(G; u, w) + 2ik(G; v)

If ik(G; u, w) > 0, then ik(G00) > ik(G), a contradiction On the other hand, if ik(G; u, w) =

0, then G + uw has the same number of k-independent sets as G, but more edges, con-tradicting the edge-maximality of G

The next natural problem is that of computing

max {i(G) : n(G) = n, ω(G) ≥ ω} Unfortunately, this question is trivial since once your graph has an ω-clique, it need not have any other edges, and so the extremal graph is Kω∪ En−ω However, the question becomes interesting if one insists that G has copies of Kω “everywhere.” For a graph G and a vertex v ∈ V (G), we define

ω(G; v) = max {|K| : K is a clique in G containing v} Theorem 3.3 If G is a graph on n vertices such that ω(G; v) ≥ ω for every v ∈ V (G), then

i(G) ≤ i K1, 1, , 1

| {z }

ω−1

,n−ω+1
= 2n−ω+1+ ω − 1

Before proving Theorem 3.3, we need to introduce quasi-threshold graphs and com-pressions that make our graphs “more quasi-threshold.”

Definition 5 A graph is quasi-threshold if, inductively, it is a single vertex, a disjoint union of two non-empty threshold graphs, or the join of a vertex and a quasi-threshold graph

Theorem 3.4 (Chv´atal, Hammer [3]) A graph is quasi-threshold if for every pair of adjacent vertices x and y, the closed neighborhoods of x and y are nested, i.e., either

N (x) ∪ {x} ⊆ N (y) ∪ {y} or N (y) ∪ {y} ⊆ N (x) ∪ {x}

Our proof of Theorem 3.3 proceeds by repeatedly applying a compression operator that produces nested closed neighborhoods

Definition 6 Let G be a graph with vertex set V (G), and suppose x ∼ y The choice

of x and y defines a natural partition of V (G) \ {x, y} into four parts: vertices which are adjacent only to x, vertices adjacent only to y, vertices adjacent to both and vertices adjacent to neither We write

Ax¯y = {v ∈ V (G) \ {x, y} : v ∼ x, v 6∼ y} ,

Axy = {v ∈ V (G) \ {x, y} : v ∼ x, v ∼ y} , and

Axy ¯ = {v ∈ V (G) \ {x, y} : v 6∼ x, v ∼ y} The compression of G from x to y, denoted Gx→y, is the graph obtained from G by deleting all edges between x and Ax¯y and adding all edges from y to Ax¯y

Lemma 3.5 If G is a graph and x ∼ y, then i(G) ≤ i(Gx→y) Also, if x and y do not have nested closed neighborhoods, we have d2(G) < d2(Gx→y), where d2(G) = P

v∈V (G)(d(v))2 Proof See [4]

Proof of Theorem 3.3 If n = ω, the result is trivial, so we assume n > ω We will select

G from the class of graphs with ω(G; v) ≥ ω for all v ∈ V (G) and i(G) maximal Among these, we select a graph with the fewest number of edges Among all such graphs, we pick one with d2(G) maximal We first note that every edge of G is in a Kω, for if e were not, then G − e would be a candidate with fewer edges

We now show that G is quasi-threshold Consider a pair x, y with x ∼ y (unless ω = 0

in which case the result is trivial) If the closed neighborhoods of x and y are not nested, then applying Lemma 3.5 we have i(Gx→y) ≥ i(G), e(Gx→y) = e(G), and d2(Gx→y) >

d2(G) Thus, if we can show that ω(Gx→y; v) ≥ ω for all v ∈ V (Gx→y) = V (G), we have contradicted the choice of G Let v ∈ V (G) with v 6= x Certainly there exists K ⊂ V (G) inducing a clique with v ∈ K If x 6∈ K, then K is also a clique in Gx→y If both x and y are in K, then K \ {x, y} ⊆ Axy, so K is also a clique in Gx→y Also, if x ∈ K but y 6∈ K, then K0 = K − x + y is a clique in Gx→y Finally, if v = x, by our earlier observation, there is a clique K00 containing the edge xy Since all of the elements of K00 are in Axy,

we have that K00 is a clique in Gx→y Thus, G is quasi-threshold

By the definition of quasi-threshold, G is either a union of non-empty quasi-threshold graphs or the join of a vertex and a quasi-threshold graph In the first case, where

G = G1∪ G2 with n(G1) = n1 and n(G2) = n2, we have

i(G) = i(G1)i(G2)

≤ (2n1 −ω+1

+ ω − 1)(2n2 −ω+1

+ ω − 1)

= 2n−2ω+2+ (ω − 1) 2n1 −ω+1+ 2n2 −ω+1
+ (ω − 1)2

≤ 2n−2ω+2 + (ω − 1) 2n−2ω+1+ 2
+ (ω − 1)2

= (ω + 1) 2n−2ω+1+ ω − 1
For ω > 2, it is straightforward to show that this final expression is strictly less than

2n−ω+1 + ω − 1 For ω = 2, we still have strict inequality unless n = 4 In this case, i(2K2) = i(K1,3) = 9

In the case where G = x ∨ G0, every vertex in G0 is contained in an (ω − 1)-clique Thus,

i(G) = i(G0) + 1 ≤ 2n−1−(ω−1)+1+ (ω − 1) − 1 + 1 = 2n−ω+1+ ω − 1,

completing the proof

4 Minimizing the number of independent sets

Another natural question to ask relates to minimizing the number of independent sets in

a graph of fixed order and size

Problem 1 Which graphs have the minimum number of independent sets amongst graphs

on n vertices and m edges?

Somewhat unexpectedly, the problem is related to the problem of maximizing the number of maximal independent sets Historically, this problem was studied in the com-plement, following a question first posed by Erd˝os and Moser, see [13] Let f (n) be the maximum number of maximal independent sets in an n-vertex graph Note that there is

no restriction on the number of edges in the graph The question was resolved indepen-dently by Miller and Muller [12] and by Moon and Moser [13] Short proofs of this result were recently given by Vatter [16] and Wood [18]

Theorem 4.1 If n ≥ 2, then

f (n) =







3n/3 if n ≡ 0 (mod 3),

4 · 3bn/3c−1 if n ≡ 1 (mod 3),

2 · 3bn/3c if n ≡ 2 (mod 3)

Further, they showed that the extremal graphs are unions of triangles, with possibly

at most one K2 or K4 It turns out that in the case where 3 divides n, the graph that maximizes the number of maximal independent sets actually minimizes the number of independent sets over graphs with n edges

The aim of this section is to solve Problem 1 when m ≤ n We begin by proving a series of lemmas that will essentially define the extremal graphs Our first lemma will be used to show that all components of an extremal graph must be unicyclic We do this

by eliminating dense components, which imply the existence of isolated vertices in this regime We extend our previous notation and write i(G; ¯x) for the number of independent sets in G not containing x Similarly, if G is a graph, and x and y are vertices of G, we write i(G; x, ¯y) for the number of independent sets containing x, but not y, and so on

We begin by proving a simple proposition

Proposition 4.2 If G is a graph, S ⊂ V (G), and x 6∈ S is a vertex of G, then

i(G; S, x) ≤ i(G; S, ¯x)

Proof Let I0 be the collection of independent sets containing S but not x, and I1 be the collection of independents sets containing S and x The map φ : I1 → I0 defined by φ(I) = I \ {x} is an injection

Armed with this proposition, we can prove a lemma that implies that graphs with at most n edges cannot have components with more than one cycle

Lemma 4.3 Suppose that G contains an edge e = xy and an isolated vertex z Then i(G) ≥ i(G − xy + xz)

Proof Let H = G − z − e and note that i(G) = 2(i(H; ¯x, ¯y) + i(H; x, ¯y) + i(H; ¯x, y)) Also

in terms of G, we note that i(G−xy+xz) = 2i(H; ¯x, ¯y)+2i(H; ¯x, y)+i(H; x, ¯y)+i(H; x, y) Thus,

i(G) − i(G − xy + xz) = 2(i(H; ¯x, ¯y) + i(H; x, ¯y) + i(H; ¯x, y))

− 2i(H; ¯x, ¯y) − 2i(H; ¯x, y) − i(H; x, ¯y) − i(H; x, y)

= i(H; x, ¯y) − i(H; x, y)

≥ 0, since i(H; x, y) ≤ i(H; x, ¯y) by Proposition 4.2

Thus, by repeatedly applying Lemma 4.3, we may assume that our extremal graphs

do not have components with more than one cycle In fact, this lemma is enough to show the result for m ≤ n/2 The extremal example consists of independent edges and isolated vertices In the regime n/2 < m ≤ n, things get a bit more interesting In the proof,

we will show that our components are unicyclic The components, then, each consist

of a cycle with trees off of the cycle We now prove a series of lemmas to narrow the possibilities for the structure of these trees The vertex labels in each lemma refer to the associated figure Recall that a cutvertex for a graph G is a vertex x such that G − x has more components than G

Lemma 4.4 Suppose G is graph containing a cutvertex x such that G − x contains a new P3-component of the type in Figure 2 Then i(G) ≥ i(G − xy + wy)

x y z w → x y z w

Figure 2: Lemma 4.4

Proof We compare G to G − xy + wy, effectively disconnecting the pendant path and replacing it with a disjoint triangle Let H = G − w − y − z We have

i(G) − i(G − xy + wy) = 3i(H; x) + 5i(H; ¯x) − 4i(H)

= i(H; ¯x) − i(H; x)

≥ 0,

by Proposition 4.2

Lemma 4.5 Suppose G is graph containing a cutvertex x such that G − x contains a new P3-component of the type in Figure 3 Then i(G) ≥ i(G − wx + yz)

x w

y

y z

→

Figure 3: Lemma 4.5

Proof Again, we compare the graphs in the lemma, and note that we are replacing a pendant tree with a disjoint triangle Let H = G − w − y − z We see that

i(G) − i(G − wx + yz) = 4i(H; x) + 5i(H; ¯x) − 4i(H)

= i(H; ¯x)

≥ 0

Lemma 4.6 Suppose G is a graph containing a cutvertex x such that G − x contains a new isolated vertex and a new P2-component as in Figure 4 Then i(G) ≥ i(G − xy −

xw + wy + wz)

of independent sets in G containing S, and similarly ik(G; S) be the number of such independent sets of size k

We